2 Indefinite Integrals

Calculus (Integrals, Series)
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jacks
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2 Indefinite Integrals

#1

Post by jacks »

(1) Evaluation of \(\displaystyle \int\frac{3x^2+1}{(x^2-1)^3}dx\)

(2) Evaluation of \(\displaystyle \int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx\) and \(\displaystyle \int\frac{\cos^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx\)
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Grigorios Kostakos
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Re: 2 Indefinite Integrals

#2

Post by Grigorios Kostakos »

Jacks here is a solution for \((1)\) : \begin{align*}
\displaystyle \int\frac{3x^2+1}{(x^2-1)^3}dx&= \int\frac{\frac{1}{2}}{(x-1)^3}-\frac{\frac{1}{2}}{(x+1)^3}dx\\
&=\frac{1}{2}\int\frac{1}{(x-1)^3}dx-\frac{1}{2}\int\frac{1}{(x+1)^3}dx\\
&=-\frac{1}{4(x-1)^2}+\frac{1}{4(x+1)^2}+c\\
&=-\frac{x}{(x^2-1)^2}+c\,.
\end{align*}
Grigorios Kostakos
jacks
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Re: 2 Indefinite Integrals

#3

Post by jacks »

I have evaluate \((a)::\) Given \(\displaystyle \int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx = \frac{1}{a^2}\int \frac{a^2\sin^2 x}{a^2\sin^2x+b^2\cos^2 x}dx\)

Divide both \(\bf{N_{r}}\) and \(\bf{D_{r}}\) by \(a^2\cos^2 x\;,\) We Get

\(\displaystyle =\frac{1}{a^2}\int\frac{\tan^2 x}{\tan^2 x+A}dx = B\int\frac{\left(\tan^2 x+A^2\right)-A^2}{\tan^2 x+A^2}dx\)

Where \(\displaystyle A^2 = \frac{b^2}{a^2}\;,\) and \(\displaystyle B=\frac{1}{a^2}\)

So Integral is \(\displaystyle =B.\int\frac{\tan^2 x+A^2}{\tan^2 x+A^2}dx-A^2.B\int\frac{1}{\tan^2 x+A^2}dx\)

Now Put \(\displaystyle \tan x=t\;,\) and \(\displaystyle dx = \frac{1}{\sec^2 x}dt = \frac{1}{1+t^2}dt\)

So Integral Convert into \(\displaystyle =B.x-B.A^2\int\frac{1}{(t^2+A^2)(1+t^2)}dt\)

So \(\displaystyle =B.x-\frac{A^2\cdot B}{\left(A^2-1\right)}\{\frac{(t^2+A^2)-(t^2+1)}{(t^2+1)(t^2+A^2)}\}dt\)

So \(\displaystyle =B\cdot x - \frac{A^2\cdot B}{\left(A^2-1\right)}\int \{\frac{1}{1+t^2}-\frac{1}{A^2+t^2}\}dt\)

So Integral is \(\displaystyle =B\cdot x -\frac{A^2\cdot B}{A^2-1}\tan^{-1}(t)-\frac{A^2\cdot B}{A^2-1}\cdot \frac{1}{A}\tan^{-1}\left(\frac{t}{A}\right)\)

Now put \(\displaystyle A=\frac{b^2}{a^2}\) and \(\displaystyle B=\frac{1}{a^2}\) and \(\displaystyle t=\tan x\;,\) We get

\(\displaystyle \int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx=\frac{x}{a^2}-\frac{b^4}{a^2\left(b^4-a^4\right)}\tan^{-1}\left(x\right)-\frac{b^4}{a^2\left(b^4-a^4\right)}\tan^{-1}\left(\frac{a^2\tan x}{b^2}\right)+\mathcal{C}\)

Now Using the identity, \(\displaystyle a^2I_{(a)}+b^2I_{(b)}=x\;,\) we get \((II)\) part of \((a)\)

Can anyone help me to calculation in \((b)\) part

Thanks
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Grigorios Kostakos
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Re: 2 Indefinite Integrals

#4

Post by Grigorios Kostakos »

Jacks, your result for \((2)\,(a)\) it is not correct: Assuming \(a\neq b\) you must have that: \[\displaystyle \int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx=\frac{1}{a^2-b^2}\Bigl({x-\frac{b}{a}\arctan\bigl({\tfrac{a}{b}\tan{x}}\bigr)}\Bigr)\]

Here is a solution for (2) α. :

\begin{align*}
\displaystyle \int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx&=\int\frac{\tan^2 x}{a^2\tan^2 x+b^2}dx\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u\,=\,\tan{x}}\\
{\frac{1}{1+u^2}\,du\,=\,dx}\\
\end{subarray}}\,\int\frac{u^2}{(a^2u^2 +b^2)(u^2+1)}du\\
&=\int{\frac{\frac{1}{a^2-b^2}}{u^2+1}-\frac{\frac{b^2}{a^2-b^2}}{a^2u^2 +b^2}du} \\
&=\frac{1}{a^2-b^2}\int{\frac{1}{u^2+1}du}-\frac{1}{a^2-b^2}\int{\frac{1}{\bigl({\frac{au}{b}}\bigr)^2+1}du} \\
&=\frac{1}{a^2-b^2}\tan{u}+c_1-\frac{1}{a^2-b^2}\frac{b}{a}\int{\frac{1}{\bigl({\frac{au}{b}}\bigr)^2+1}d\bigl({\tfrac{au}{b}}\bigr)} \\
&=\frac{1}{a^2-b^2}\arctan{u}-\frac{1}{a^2-b^2}\frac{b}{a}\arctan\bigl({\tfrac{au}{b}}\bigr)+c \\
&\stackrel{u\,=\,\tan{x}}{=\!=\!=\!=\!=}\frac{1}{a^2-b^2}\Bigl({\arctan(\tan{x})-\frac{b}{a}\arctan\bigl({\tfrac{a}{b}\tan{x}}\bigr)}\Bigr)+c \\
&=\frac{1}{a^2-b^2}\Bigl({x-\frac{b}{a}\arctan\bigl({\tfrac{a}{b}\tan{x}}\bigr)}\Bigr)+c\,.
\end{align*}
Grigorios Kostakos
jacks
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Re: 2 Indefinite Integrals

#5

Post by jacks »

Grigorios Kostakos wrote:Jacks, your result for \((2)\,(a)\) it is not correct: Assuming \(a\neq b\) you must have that: \[\displaystyle \int\frac{\sin^2 x}{a^2\sin^2 x+b^2\cos^2 x}dx=\frac{1}{a^2-b^2}\Bigl({x-\frac{b}{a}\arctan\bigl({\tfrac{a}{b}\tan{x}}\bigr)}\Bigr)\]
Thanking You Grigorios Kostakos, Got my mistake.
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